Mrs. Smith is trying to contact her husband, John Smith, but she forgot the secret phone number!
The only thing Mrs. Smith remembered was that any permutation of nn can be a secret phone number. Only those permutations that minimize secret value might be the phone of her husband.
The sequence of nn integers is called a permutation if it contains all integers from 11 to nn exactly once.
The secret value of a phone number is defined as the sum of the length of the longest increasing subsequence (LIS) and length of the longest decreasing subsequence (LDS).
A subsequence ai1,ai2,…,aikai1,ai2,…,aik where 1≤i1<i2<…<ik≤n1≤i1<i2<…<ik≤n is called increasing if ai1<ai2<ai3<…<aikai1<ai2<ai3<…<aik. If ai1>ai2>ai3>…>aikai1>ai2>ai3>…>aik, a subsequence is called decreasing. An increasing/decreasing subsequence is called longest if it has maximum length among all increasing/decreasing subsequences.
For example, if there is a permutation [6,4,1,7,2,3,5][6,4,1,7,2,3,5], LIS of this permutation will be [1,2,3,5][1,2,3,5], so the length of LIS is equal to 44. LDScan be [6,4,1][6,4,1], [6,4,2][6,4,2], or [6,4,3][6,4,3], so the length of LDS is 33.
Note, the lengths of LIS and LDS can be different.
So please help Mrs. Smith to find a permutation that gives a minimum sum of lengths of LIS and LDS.
The only line contains one integer nn (1≤n≤1051≤n≤105) — the length of permutation that you need to build.
Print a permutation that gives a minimum sum of lengths of LIS and LDS.
If there are multiple answers, print any.
4
3 4 1 2
2
2 1
In the first sample, you can build a permutation [3,4,1,2][3,4,1,2]. LIS is [3,4][3,4] (or [1,2][1,2]), so the length of LIS is equal to 22. LDS can be ony of [3,1][3,1], [4,2][4,2], [3,2][3,2], or [4,1][4,1]. The length of LDS is also equal to 22. The sum is equal to 44. Note that [3,4,1,2][3,4,1,2] is not the only permutation that is valid.
In the second sample, you can build a permutation [2,1][2,1]. LIS is [1][1] (or [2][2]), so the length of LIS is equal to 11. LDS is [2,1][2,1], so the length of LDS is equal to 22. The sum is equal to 33. Note that permutation [1,2][1,2] is also valid.
题意 : 给一个数 n, 求一个长度为 n 的排列,使得该排列的 lis长度 + lds 长度 最小。
思路:取 ans = x + y (x 表示 lis 长度,y 表示 lds 长度) 那么 先将 n 个数字分成 p 块,使每个块中单调递增序列,令 p 块中的数字都为相邻的数。例如 n=9 时,令 块一:123、块二:456、块三:789,要是 x + y最小,我们可以得出当排列为块三、块二、块一时最小,即 ans = n/p+p。由数学不等式得 : n/p+p>=2*sqrt(n),即p==sqrt(n)时,ans取最小值 2*sqrt(n) 。由特殊到一般的归纳法,我们即可推出规律。
代码如下:
#include <bits/stdc++.h> using namespace std; const int maxn=1e5+10; int n; int main(){ scanf("%d",&n); int m=sqrt(n); for (int i=0; i<n; i++) printf("%d ",max(0,n-m*(i/m+1))+i%m+1); return 0; }